#-*- coding: UTF-8 -*- # Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = None#Method1class Solution(object):    def mergeTwoLists(self, l1, l2):        """  …
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 # Definition for singly-link…
1.题目 21. Merge Two Sorted Lists Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4-&g…
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:合并,有序链表,递归,迭代,题解,leetcode, 力扣,Python, C++, Java 目录 题目描述 题目大意 解题方法 迭代 Python解法 C++解法 Java解法 递归 日期 题目地址:https://leetcode.com/problems/merge-two-sorted-lists/ 题目描述 Merge two sorted…
一天一道LeetCode系列 (一)题目 Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. (二)解题 这题是剑指offer上的老题了,剑指上面用的是递归,我写了个非递归的版本. /** * Definition for singly-linked list. *…
题目: Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 思路:比较每个列表的第一个元素. 合并小的添加到列表中. 最后,当其中一个是空的,只需将它附加到合并后的列表,因为它已经排序. /** * Definition for singly-linked list.…
题目描述: Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 解题分析: 再基础不过的题了,直接看代码吧^-^ 具体代码: /** * Definition for singly-linked list. * public class ListNode { * in…
算法和数据结构这东西,真的是需要常用常练.这道看似简单的链表合并题,难了我好几个小时,最后还是上网搜索了一种不错算法.后期复习完链表的知识我会将我自己的实现代理贴上. 这个算法巧就巧在用了递归的思想,按照常规方法也能求得,但是就未免太复杂了. Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the fir…
#-*- coding: UTF-8 -*-class Solution(object):    def merge(self, nums1, m, nums2, n):        """        :type nums1: List[int]        :type m: int        :type nums2: List[int]        :type n: int        :rtype: void Do not return anything,…
21. Merge Two Sorted Lists[easy] Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 解法一: /** * Definition for singly-linked list. * struct ListNode { * int val…
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:合并,链表,单链表,题解,leetcode, 力扣,Python, C++, Java 题目地址: https://leetcode.com/problems/merge-k-sorted-lists/description/ 题目描述: Merge k sorted linked lists and return it as one sorted li…
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 21: Merge Two Sorted Listshttps://oj.leetcode.com/problems/merge-two-sorted-lists/ Merge two sorted linked lists and return it as a new list.The new list should be made by splicing together…
21. Merge Two Sorted Lists Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists 需要排序!!! [2,4][1] 输出1 2 4 而不是2 4 1  递归版!! class Solution { public ListNode mergeTwo…
21.Merge Two Sorted Lists 初始化一个指针作为开头,然后返回这个指针的next class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* dummy = ); ListNode* p = dummy; while(l1 && l2){ if(l1->val <= l2->val){ p->next = l1; p = p-&…
一.题目说明 这个题目是21. Merge Two Sorted Lists,归并2个已排序的列表.难度是Easy! 二.我的解答 既然是简单的题目,应该一次搞定.确实1次就搞定了,但是性能太差: Runtime: 20 ms, faster than 8.74% of C++ online submissions for Merge Two Sorted Lists. Memory Usage: 9.4 MB, less than 5.74% of C++ online submissions…
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 这道混合插入有序链表和我之前那篇混合插入有序数组非常的相…
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 和88. Merge Sorted Array类似,数据…
21. Merge Two Sorted Lists Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 代…
合并链表 Runtime: 4 ms, faster than 100.00% of C++ online submissions for Merge Two Sorted Lists. class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { //1 2 4 . 1 3 4 ListNode *res = ); ListNode *cur = res; while (l1 != NULL &&am…
题目描述(easy) Merge Two Sorted Lists Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4-…
1. Merge Two Sorted Lists 题目链接 题目要求:  Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 这道题目题意是要将两个有序的链表合并为一个有序链表.为了编程方便,在程序中引入dummy节点.具体程序如下: /** * Definitio…
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 思路.首先判断是否有空链表,如果有,则直接返回另一个链表,如果没有,则开始比较两个链表的当前节点,返回较小的元素作为前驱,并且指针向后移动一位,再进行比较,如此循环,知道一个链表的next指向NULL,将另一个链表的…
一天一道LeetCode系列 (一)题目 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. (二)解题 合并K个已拍好序的链表.剑指上有合并两个已排好序的链表的算法,那么K个数,我们可以采用归并排序的思想,不过合并函数可能需要修改一下,换成合并两个已排好序的链表的方法.代码如下: /** * Definition for singly-linked…
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题解: 归并思想. Solution 1 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {…
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 题解: 简单的链表遍历,还可用递归做. Solution…
题目: Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 没事来做做题,该题目是说两个排序好的链表组合起来,依然是排序好的,即链表的值从小到大. 代码: 于是乎,新建一个链表,next用两个链表当前位置去比较,谁的小就放谁.当一个链表放完之后,说明另外一个链表剩下的…
#-*- coding: UTF-8 -*- #既然不能使用加法和减法,那么就用位操作.下面以计算5+4的例子说明如何用位操作实现加法:#1. 用二进制表示两个加数,a=5=0101,b=4=0100:#2. 用and(&)操作得到所有位上的进位carry=0100;#3. 用xor(^)操作找到a和b不同的位,赋值给a,a=0001:#4. 将进位carry左移一位,赋值给b,b=1000:#5. 循环直到进位carry为0,此时得到a=1001,即最后的sum.#!!!!!!关于负数的运算.…
题目地址: https://oj.leetcode.com/problems/merge-k-sorted-lists/ 题目内容: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ Merge k sorted linked lists and return it a…
题目 Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 翻译 合并两个有序的链表 Hints Related Topics: LinkedList 参考 归并排序-维基百科,可以递归也可以迭代,基本的链表操作 代码 Java /** * Definition for…
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 题目标签:Linked List 题目给了我们两个lists,让我们有序的合并两个 lists. 这题利用递归可以从list 的最后开始向前链接nodes,代码很简洁,清楚. Java Solution: Runti…